I know Cramer's rule for $2$ and $3$ unknowns.
Is there a Cramer's Rule for $n≥4$? (where $n$ is the number of unknowns)
I'm not able to find it on the net.
What is the idea behind the Cramer's rule? How can I write the Cramer's rule for $n$ equations with $n$ unknowns without memorising any sequence?
EDIT : I did check the Wikipedia and I couldn't understand anything :(
I'm not sure if I fully understand the question, but if you want some compact expression for Cramer's rule, probably I can help and provide you with one -- the only thing it requires you to memorize is how to calculate determinants.
SOLUTION
Consider the following equation $$ \underbrace{ \begin{pmatrix} \begin{matrix} a_{1,1} \\ a_{2,1} \\ \cdots \\ a_{n,1} \\ \end{matrix} & % \begin{matrix} a_{1,2} \\ a_{2,2} \\ \cdots \\ a_{n,2} \\ \end{matrix} & % \begin{matrix} \cdots \\ \cdots \\ \cdots \\ \cdots \\ \end{matrix} & % \begin{matrix} a_{1,n} \\ a_{2,n} \\ \cdots \\ a_{n,n} \\ \end{matrix} \end{pmatrix} }_{\hat{A}} \underbrace{ \begin{pmatrix} x_{1} \vphantom{x_{1}^{(1)}} \\ x_{2} \vphantom{x_{1}^{(1)}} \\ \cdots \\ x_{n} \vphantom{x_{1}^{(1)}} \\ \end{pmatrix} }_{\vec{x}} = \underbrace{ \begin{pmatrix} b_{1} \vphantom{x_{1}^{(1)}} \\ b_{2} \vphantom{x_{1}^{(1)}} \\ \cdots \\ b_{n} \vphantom{x_{1}^{(1)}} \\ \end{pmatrix} }_{\vec{b}}, $$ where $\hat{A}$ and $\vec{b}$ are given and we aim to find $\vec{x}$. Solution may be written as $$ \begin{pmatrix} x_{1} \vphantom{x_{1}^{(1)}} \\ x_{2} \vphantom{x_{1}^{(1)}} \\ \cdots \\ x_{n} \vphantom{x_{1}^{(1)}} \\ \end{pmatrix} = (-1) \frac{ \det \begin{pmatrix} 0 & \vec{e}_{1} & \vec{e}_{2} & \dots & \vec{e}_{n} \\ \begin{matrix} b_{1} \vphantom{x_{1}^{(1)}} \\ b_{2} \vphantom{x_{1}^{(1)}} \\ \cdots \\ b_{n} \vphantom{x_{1}^{(1)}} \\ \end{matrix} & % \begin{matrix} a_{1,1} \\ a_{2,1} \\ \cdots \\ a_{n,1} \\ \end{matrix} & % \begin{matrix} a_{1,2} \\ a_{2,2} \\ \cdots \\ a_{n,2} \\ \end{matrix} & % \begin{matrix} \cdots \\ \cdots \\ \cdots \\ \cdots \\ \end{matrix} & % \begin{matrix} a_{1,n} \\ a_{2,n} \\ \cdots \\ a_{n,n} \\ \end{matrix} \end{pmatrix} }{ \det \begin{pmatrix} \begin{matrix} a_{1,1} \\ a_{2,1} \\ \cdots \\ a_{n,1} \\ \end{matrix} & % \begin{matrix} a_{1,2} \\ a_{2,2} \\ \cdots \\ a_{n,2} \\ \end{matrix} & % \begin{matrix} \cdots \\ \cdots \\ \cdots \\ \cdots \\ \end{matrix} & % \begin{matrix} a_{1,n} \\ a_{2,n} \\ \cdots \\ a_{n,n} \\ \end{matrix} \end{pmatrix} }, $$ where $\vec{e}_1$, $\dots$, $\vec{e}_n$ are unitary vetors along coordinate axes.
STRUCTURE OF THE SOLUTION
I'll rewrite the equation once more to show how it's built $$ \begin{pmatrix} x_{1} \vphantom{x_{1}^{(1)}} \\ x_{2} \vphantom{x_{1}^{(1)}} \\ \cdots \\ x_{n} \vphantom{x_{1}^{(1)}} \\ \end{pmatrix} = (-1) \frac{ \det \begin{pmatrix} 0 & \boxed{\begin{matrix} \vec{e}_{1} & \vec{e}_{2} & \dots & \vec{e}_{n} \end{matrix}} \\ \boxed{ \begin{matrix} b_{1} \vphantom{x_{1}^{(1)}} \\ b_{2} \vphantom{x_{1}^{(1)}} \\ \cdots \\ b_{n} \vphantom{x_{1}^{(1)}} \\ \end{matrix}} & % \boxed{ \begin{matrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n}\\ a_{2,1} & a_{2,2} & \cdots & a_{2,n}\\ \cdots & \cdots & \cdots & \cdots\\ a_{n,1} & a_{n,2} & \cdots & a_{n,n}\\ \end{matrix} } \end{pmatrix} }{ \det \begin{pmatrix} \boxed{ \begin{matrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n}\\ a_{2,1} & a_{2,2} & \cdots & a_{2,n}\\ \cdots & \cdots & \cdots & \cdots\\ a_{n,1} & a_{n,2} & \cdots & a_{n,n}\\ \end{matrix} } \end{pmatrix} } = (-1) \frac{ \det \begin{pmatrix} 0 & \boxed{\begin{matrix} \vec{e}_{1} \dots & \vec{e}_{n} \end{matrix}} \\ \boxed{\vec{b}\\} & \boxed{\hspace{1cm}\hat{A}\hspace{1cm}\\} \end{pmatrix} }{ \det \begin{pmatrix} \boxed{\hat{A}} \end{pmatrix} } $$
EXAMPLE
Consider following equations $$ \left\{ \begin{aligned} -2 x_1 + 3 x_2 + x_3 &= 2, \\ - x_1 - x_2 + 4 x_3 &= 1, \\ x_1 + x_2 + x_3 &= 1, \end{aligned} \right. $$ or in matrix form $$ \begin{pmatrix} -2 & 3 & 1 \\ -1 & -1 & 4 \\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 1 \\ \end{pmatrix}. $$
Solution is written as $$ \vec{x} = (-1) \frac{ \det \begin{pmatrix} 0 & \vec{e}_1 & \vec{e}_2 & \vec{e}_3 \\ 2 & -2 & 3 & 1 \\ 1 & -1 & -1 & 4 \\ 1 & 1 & 1 & 1 \end{pmatrix} }{ \det \begin{pmatrix} -2 & 3 & 1 \\ -1 & -1 & 4 \\ 1 & 1 & 1 \end{pmatrix} } = \frac{1}{25} \vec{e}_1 + \frac{14}{25} \vec{e}_2 + \frac{2}{5}\vec{e}_3 =\begin{pmatrix} \frac{1}{25} \\ \frac{14}{25} \\ \frac{2}{5} \\ \end{pmatrix}, $$ thus $x_1=1/25$, $x_2=14/25$, and $x_3=2/5$.
REFERENCES
To read more on the theory everything relies on, you may want to check "Beginner's guide to mapping simplexes affinely" that is written by authors of the formula. Besides they have "Workbook on mapping simplexes affinely" that contains different examples with similar determinants.